# Finding the turning points of $f(x)=\left(x-a+\frac1{ax}\right)^a-\left(\frac1x-\frac1a+ax\right)^x$

I've just come across this function when playing with the Desmos graphing calculator and it seems that it has turning points for many values of $a$.

So I pose the following problem:

Given $a \in \mathbb{R}-\{0\}$, find $x$ such that $\dfrac{dy}{dx}=0$ where $y=\left(x-a+\dfrac1{ax}\right)^a-\left(\dfrac1x-\dfrac1a+ax\right)^x$

As in most maxima/minima problems, we first (implicitly) differentiate it and set to $0$ to give $$\boxed{\small\dfrac{a(ax^2-1)}{x(ax^2-a^2x+1)}\left(\dfrac{ax^2-a^2x+1}{ax}\right)^a=\left(\ln\left(\dfrac{a^2x^2-x+a}{ax}\right)+\dfrac{a(ax^2-1)}{a^2x^2-x+a}\right)\left(\dfrac{a^2x^2-x+a}{ax}\right)^x} \tag{1}$$ I have no idea how to continue from here. I thought about taking logarithms, but it appears to me that the double $\ln$ in the term $\dfrac{a^2x^2-x+a}{ax}$ would only make the equation worse.

(For the simplest case when $a=1$, the problem is easy: $x=1$ and it is a point of inflexion).

Let's try setting each of the terms to $0$:

## Case $1$: $\left(\frac{a^2x^2-x+a}{ax}\right)^x=0$

$\hspace{1cm}$ This is only possible when the fraction is zero; that is, solving $a^2x^2-x+a=0$ to get $$x=\frac{1\pm\sqrt{1-4a^3}}{2a^2}$$

## Case $2$: $\left(\frac{ax^2-a^2x+1}{ax}\right)^a=0$

$\hspace{1cm}$ This gives \begin{align}ax^2-a^2x+1=0&\implies a^2x^2+a=a^3x\\&\implies\left(\dfrac{a^2x^2-x+a}{ax}\right)^x=\left(\dfrac{a^3x-x}{ax}\right)^x=\left(\dfrac{a^3-1}{a}\right)^x=0\end{align}

$\hspace{1cm}$ so for equality between LHS and RHS, we must have $a=1$. However, the equation

$\hspace{1cm}$ $ax^2-a^2x+1=0$ has no real solutions for such $a$; hence we reach a contradiction.

## Case $3$: $\frac{a(ax^2-1)}{x(ax^2-a^2x+1)}=0$

$\hspace{1cm}$ We have $x=\pm \dfrac1a$. Now LHS is $0$, and $$\left(\dfrac{a^2x^2-x+a}{ax}\right)^x=\left(1-\dfrac1a+a\right)^{\frac1a} \neq 0$$

$\hspace{1cm}$ for $a \in \mathbb{R} - \{\phi\}$, where $\phi$ is the golden ratio.

$\hspace{1cm}$ Suppose that $a = \phi$. Then $x$ is forced to be $-\dfrac1a=-\dfrac2{1+\sqrt5}$, since $ax^2-a^2x+1=0$

$\hspace{1cm}$ (undefined) when $x=\dfrac 1a$. This is impossible, since $y$ is only defined when $x>0$ for this

$\hspace{1cm}$ value of $a$!

## Case $4$: $\ln\left(\frac{a^2x^2-x+a}{ax}\right)+\frac{a(ax^2-1)}{a^2x^2-x+a}=0$

$\hspace{1cm}$ This is impossible from cases $1$ and $3$.

UPDATE: I have provided a partial answer to my question, now with $x$ removed from it.

Any hints on how to solve $(3)=(4)$ are welcome.

• One worthwhile thing to note is you have the product of two functions so each separately may be set to zero much like one might with a quadratic, but somewhere there should be a power of $a-1$ for the power rule which appears to be missing Commented Dec 2, 2017 at 21:51
• The power of $a-1$ isn't missing: $\dfrac{(ax^2-a^2x+1)^a}{ax^2-a^2x+1}=(ax^2-a^2x+1)^{a-1}$. I used implicit differentiation to find the derivatives of each term in $y$. Commented Dec 3, 2017 at 10:20
• I think you can express $x$ as a serie expansion with the Lagrange inversion theorem mathworld.wolfram.com/LagrangeInversionTheorem.html . Commented Dec 6, 2017 at 11:26
• This is the case in which you have to consider two separate functions and find max and min of each and sum the results. This is practical and easy. Commented Dec 21, 2017 at 8:26
• @TheSimpliFire Did you find the domain of (1)?
– user480281
Commented Dec 30, 2017 at 22:20

A note about Case $1$ with $\,a:= 2^{-\frac{2}{3}}\,$ and $\,\displaystyle x\to 2^{\frac{1}{3}}\,$ .

Left side $\,=0\,$ for $\,\displaystyle x=2^{\frac{1}{3}}\,$ because of $\, ax^2-1=0\,$ and $\,\displaystyle ax^2-a^2x+1=\frac{3}{2}\ne 0\,$ .

Right side $\,=0\,$ for $\,\displaystyle x\to 2^{\frac{1}{3}}>1\,$:

$\,\displaystyle \lim_{z\to +0 \\x>0} z^x\ln z=0\,$ and therefore $\,\displaystyle \left(\frac{a^2x^2-x+a}{ax}\right)^x \ln \frac{a^2x^2-x+a}{ax} \to 0\,$ for $\,\displaystyle x\to 2^{\frac{1}{3}}>0\,$

$\,\displaystyle \left(\frac{a^2x^2-x+a}{ax}\right)^x \frac{a(ax^2-1)}{a^2x^2-x+a} = \left(a^2x^2-x+a\right)^{x-1} (ax^2-1) x^{-x} a^{1-x} = 0\,$

for $\,\displaystyle x=2^{\frac{1}{3}}\,$ because of $\,x-1>0\,$ and $\,\displaystyle a^2x^2-x+a=0\,$ and $\, ax^2-1=0\,$ .

If you like to work with recursions for e.g. $\,a>0,\,a\neq 1\,$

it can make sense to choose $\,\displaystyle z:=x+\frac{1}{ax}\,$ so that you get

$\displaystyle x=f_{1,2}(z):=\frac{z}{2}\pm\sqrt{(\frac{z}{2})^2-\frac{1}{a}}\,$ and $\,\displaystyle y=(z-a)^a-\left(az-\frac{1}{a}\right)^x\,$ .

We get $\,\displaystyle \frac{dy}{dz}=(z-a)^{a-1} - \left(az-\frac{1}{a}\right)^x \left(\frac{x}{z-\frac{1}{a^2}}+\frac{1}{2-\frac{z}{x}}\ln\left(az-\frac{1}{a}\right)\right)$

and a possible recursion with $\displaystyle z_0>\max\left(a;\frac{2}{\sqrt{a}};\frac{1}{a^2}\right)\,$ could be

$\displaystyle z_{n+1}=a+\left(\left(az_n-\frac{1}{a}\right)^{f(z_n)} \left(\frac{f(z_n)}{z_n-\frac{1}{a^2}}+\frac{1}{2-\frac{z_n}{f(z_n)}}\ln\left(az_n-\frac{1}{a}\right)\right)\right)^{\frac{1}{a-1}}\,$

with $\,f(z)\in\{f_1(z);f_2(z)\}\,$ .

For time reasons I haven't checked this recursion, sorry. But it's a try to simplify the calculations.

Note:

Because of $\,\displaystyle \frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}\,$ with $\,\displaystyle \frac{dy}{dx}:=0\,$ we have $\,\displaystyle \frac{dy}{dz}=0\,$ or $\,\displaystyle \frac{dz}{dx}=0\,$ .

$\displaystyle \frac{dz}{dx}=0\,$ means $\,ax^2-1=0\,$ which leads directly to my comment about Case $1$ .

This strengthens the claim that the substitution $\,\displaystyle z:=x+\frac{1}{ax}\,$ makes sense.

• Ah, thanks. I didn't think about the fraction being zero! Will edit. Commented May 3, 2018 at 13:49
• @TheSimpliFire : Thanks for the bounty, very kind of you! Commented May 5, 2018 at 9:10

I've made some progress on the question, but I'm nowhere close to solving it.

For ease of reading, I will repeat the boxed equation below: $$\small\dfrac{a(ax^2-1)}{x(ax^2-a^2x+1)}\left(\dfrac{ax^2-a^2x+1}{ax}\right)^a=\left(\ln\left(\dfrac{a^2x^2-x+a}{ax}\right)+\dfrac{a(ax^2-1)}{a^2x^2-x+a}\right)\left(\dfrac{a^2x^2-x+a}{ax}\right)^x$$

At the end of my post above, there was a suggestion of setting $$\left(\dfrac{ax^2-a^2x+1}{ax}\right)^a=k$$ in the hope that we can find $k$ in terms of $a$, since only this has simple solutions for $x$.

The consequences of this are as follows: $$ax^2+1=a(a+\sqrt[a]k)x\tag{1}$$$$\small x=\frac{a+\sqrt[a]k}2\pm\frac{\sqrt{a^2(a+\sqrt[a]k)^2-4a}}{2a}\implies x^2=\frac{a(a+\sqrt[a]k)^2\pm(a+\sqrt[a]k)\sqrt{a^2(a+\sqrt[a]k)^2-4a}-2}{2a}\tag{2}$$

This means that using $(1)$, $$\frac{a(ax^2-1)}{x(ax^2-a^2x+1)}=\frac{a(ax^2-1)}{x(ax^2+1)-a^2x^2}=\frac{ax^2-1}{\sqrt[a]kx^2}=\boxed{\frac a{\sqrt[a]k}-\frac1{\sqrt[a]kx^2}}$$

We now deal with the natural logarithm term on the RHS, using $(1)$. Notice that $$\ln\left(\frac{a^2x^2-x+a}{ax}\right)=\ln\left(\frac{a(ax^2+1)}{ax}-\frac1a\right)=\boxed{\ln\left(a(a+\sqrt[a]k)-\frac1a\right)}$$ Hooray, we've got rid of the $x$! Similarly, we have $$\left(\frac{a^2x^2-x+a}{ax}\right)^x=\boxed{\left(a(a+\sqrt[a]k)-\frac1a\right)^x}$$

The last term next to the logarithmic term is a pain in the neck. However, $(1)$ can still be used. $$\frac{a(ax^2-1)}{a^2x^2-x+a} = \frac{a(ax^2+1)-2a}{a(ax^2+1)-x}=\boxed{\frac{a^2(a+\sqrt[a]k)x-2a}{a^2(a+\sqrt[a]k)-x}}$$

Let's combine all of the boxed terms to see what we currently have: $$k\left(\frac a{\sqrt[a]k}-\frac1{\sqrt[a]kx^2}\right)=\left(\ln\left(a(a+\sqrt[a]k)-\frac1a\right)+\frac{a^2(a+\sqrt[a]k)x-2a}{a^2(a+\sqrt[a]k)-x}\right)\left(a(a+\sqrt[a]k)-\frac1a\right)^x$$

Now we use $(2)$. The steps to the final result are horrible, so I will just give it here. Note that when I took logarithms in the RHS, I used the fact that $\ln(ab^c)=\ln a + c\ln b$. Just hope I haven't made any mistakes: $$\text{LHS}=\frac k{\sqrt[a]k}\left(1-\frac{4a-2}{a^2(a+\sqrt[a]k)^2\pm(a+\sqrt[a]k)\sqrt{a^2(a+\sqrt[a]k)^2-4a}-2}\right)\tag{3}$$ and \begin{align}\ln(\text{RHS})&=\frac{a(a+\sqrt[a]k)\pm\sqrt{a^2(a+\sqrt[a]k)^2-4a}}{2a}\ln\left(a(a+\sqrt[a]k)-\frac1a\right)\tag{4}\\\small&+\ln\left(\ln\left(a(a+\sqrt[a]k)-\frac1a\right)+\frac{a^3(a+\sqrt[a]k)^2\pm a^2(a+\sqrt[a]k)\sqrt{a^2(a+\sqrt[a]k)^2-4a}-4}{2a^3(a+\sqrt[a]k)-(a+\sqrt[a]k)\pm\sqrt{a^2(a+\sqrt[a]k)^2-4a}}\right)\end{align} Is it even possible now to rearrange this to make $k$ the subject?

Here is a visualisation of the equation $\text{LHS}=\text{RHS}$.

This is not an answer, just a comment.

Your $f(a,x)=\left(x-a+\dfrac1{ax}\right)^a-\left(\dfrac1x-\dfrac1a+ax\right)^x$ gives us for different $a$ functions of $x$ that probably have different domains.

So, when you say "Given $a \in \mathbb{R}$" that is not at all as precise as it could be. It would be more precise if we could for every $a\in \mathbb R$ calculate the domain of $f(a,x)$, because, when you differentiated this function in order to find critical point(s), are you sure that they are in the domain of $f$, even if you succeed in calculating them?

It is natural to think that as $a$ vary continuously that then the domains of $f$ will vary continuously but it is a question of what happens for $a=0$, that is, does $\dfrac {1}{f(a,x)}$ has limit when $a \to 0$, or is discontinuous (in the sense of domains)?

I am of the opinion that finding solutions to your (1) should not be done before first calculating for which $a \in \mathbb R$ your $f$ is well-defined function of $x$, and what is the domain of $f$ for every $a \in \mathbb R$.